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bixtya [17]
3 years ago
10

List the first 20 elements chemistry

Chemistry
1 answer:
timama [110]3 years ago
8 0

Answer:

h- Hydrogen

He - Helium

Li - Lithium

Be - Beryllium

B - Boron

C - Carbon

N - Nitrogen

O - Oxygen

Explanation:

Happy to help:)

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Someone help please , I’ll give brain . Question 10 !
Rudik [331]

Answer:it’s c

Explanation:

I just know it is

4 0
3 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

3 0
3 years ago
A substance formed when two or more elements combine and lose their distinct properties is
hram777 [196]
Is a compound that went through a chemicAl change
7 0
2 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What
coldgirl [10]

Answer:

\rm_{90}^{231}\text{Th}

Explanation:

The unbalanced nuclear equation is

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + X

Let's write X as a nuclear symbol.

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} + _{Z}^{A}\text{X}

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

 92 = 2 + Z , so  Z =   92 - 2 =  90

And your nuclear equation becomes

\rm _{92}^{235}\text{U} \longrightarrow \,  _{2}^{4}\text{He} +\, _{90}^{231}\text{X}

Element 90 is thorium, so  

\rm X = _{90}^{231}\text{Th}

7 0
3 years ago
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